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Concerning Laurent series

  1. Dec 6, 2008 #1
    I'm puzzled by one thing concerning Laurent series.

    If I have a series, for example f(z) =(z*sinz)/(2z-1) and I'm supposed to make a laurent series of f about the point z=1/2.

    Now, what would the inner and outer radius of convergence be?

    I would say that since z=1/2 is a pole, the inner radius is zero and the outer radius infinite?

    If not, then how can I see the radius of convergence of a Laurent series?
  2. jcsd
  3. Dec 6, 2008 #2
    Result is correct, and the first conclusion too. Inner radius can be greater than zero only if Laurent series don't terminate at negative direction. Now the smallest order term in the Laurent series is of order -1, so inner radius is zero.

    The outer radius is infinite, but you cannot conclude it from the fact that z=1/2 is a pole. Instead it is seen most easily from the fact, that the series of sine have the factorials 1/n! which make series convergent everywhere.

    Most of the time it is best to try to see them by using previous knowledge of already known functions, but when all else fails, the outer and inner radius, [itex]\rho_O[/itex] and [itex]\rho_I[/itex], of a Laurent series

    \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n

    are given by the formulas

    \rho_O = \frac{1}{\underset{n\to\infty}{\textrm{lim sup}}\; \sqrt[n]{|a_n|}}

    \rho_I = \underset{n\to\infty}{\textrm{lim sup}}\; \sqrt[n]{|a_{-n}|}
  4. Dec 6, 2008 #3
    Then for example exp(1/z) has infinite negative terms and so has an inner radius higher than zero?

    Another thing. If I wanted to find the residue of the pole(z=1/2) by writing the function as an infinite series, how would I do that? I would have to write it in terms of (z-1/2) right?

    I tried doing it for sine about z=1/2 but the the denominator cancelled and I did not gera a -1 order term in the series.
  5. Dec 6, 2008 #4

    This claim is right: If Laurent series has only finite amount of non-zero terms with negative exponent, then the inner radius is zero.

    This claim is wrong: If Laurent series has infinite amount of non-zero terms with negative exponent, then the inner radius is greater than zero.

    These first claim does not imply the second.

    (edit: and my original comment
    is equivalent with the first claim)

    Yes. You need to solve the coefficient [itex]a_{-1}[/itex] from the series

    \frac{z\sin(z)}{2z-1} \;=\; \frac{a_{-1}}{z-\frac{1}{2}} \;+\; a_0 \;+\; a_1(z-\frac{1}{2}) \;+\; a_2(z-\frac{1}{2})^2 \;+\; \cdots

    I believe you made a mistake. [itex]\frac{1}{2}\sin(\frac{1}{2})\neq 0[/itex], so the -1 order term should be there.
  6. Dec 8, 2008 #5
    I don´t understand how that leads to the a-1 term. Is it because then the function would have a rectifiable singularity, that is the limit would exist as z goes to 1/2?

    These things are confusing me a bit. For example I don´t get a a-1 term in a function like f(z) = 1/(z-i) if I expand it as a geometric series. Well, I get an a-1 term but an infinite number of negative powers, and that, by definition, should not give me a pole, but something called an essential singularity.

    But I KNOW that z=i is a pole, since the function goes to infinity as z->i
  7. Dec 8, 2008 #6
    I'm not sure what rectifiable singularity means. If [itex]f(z_0)=0[/itex], then [tex]\frac{f(z)}{z-z_0}[/tex] has a removable singularity at [itex]z_0[/itex], at least. It sounds like you understood this already, but merely weren't sure?

    This function already is in the form of Laurent series. It is

    \frac{1}{z-i} = \sum_{n=-\infty}^{\infty} a_n (z-i)^n

    with coefficients [itex]a_{-1} = 1[/itex] and [itex]a_n = 0[/itex] for [itex]n\neq -1[/itex].

    If you did this:

    \frac{1}{z-i} = \frac{i}{1 + iz} = i\sum_{n=0}^{\infty} (-iz)^n = \sum_{n=0}^{\infty} (-1)^n i^{n+1} z^n

    then you have expanded the function as Laurent series around the point [itex]z=0[/itex], and not around the point [itex]z=i[/itex]. In this case the inner radius of convergence is 0, and outer radius is 1. There does not exist [itex]z^{-1}[/itex] term with non-zero coefficient, because the original function doesn't have singularity at [itex]z=0[/itex]. The singularity is at [itex]z=i[/itex], and it has the effect that the (open) domain of convergence cannot be larger than [itex]\{z\in\mathbb{C}\;|\; |z|<1\}[/itex], because a larger ball would contain the [itex]z=i[/itex]. The singularity prevents the domain of convergence from being larger.

    Now I'm guessing, but do you mean series like this?

    \frac{1}{z-i} = \frac{1}{z} \frac{1}{1 - \frac{i}{z}} = \frac{1}{z} \sum_{n=0}^{\infty} \Big(\frac{i}{z}\Big)^n = \sum_{n=-\infty}^{-1} (-i)^{n+1} z^n

    Here the function is written as Laurent series around the point [itex]z=0[/itex] so that the (open) domain of convergence is [itex]\{z\in\mathbb{C}\;|\; |z|>1\}[/itex]. That means the inner radius of convergence is 1, and the outer radius is [itex]\infty[/itex]. The original function doesn't have essential singularity at [itex]z=0[/itex] despite these negative powers in this expansion, because now [itex]z=0[/itex] is outside (and not even on the boundary) the domain of convergence. To see what happens at the [itex]z=0[/itex], you need to use the Laurent series which converge for [itex]|z|<1[/itex].
    Last edited: Dec 8, 2008
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